Equivariant singular cohomology - MathOverflow

Equivariant singular cohomology

2009-12-23T02:30:32Z

One can define the $G$-equivariant cohomology of a space $X$ as being the ordinary singular cohomology of $X \times_G EG$ --- I think this is due to Borel? (See e.g. section 2 of these notes)

Alternatively if $X$ is a manifold, we also have $G$-equivariant de Rham cohomology, defined in terms of $G$-equivariant differential forms --- I think this is due to Cartan? (See e.g. section 3 of loc. cit.)

I suspect this is extremely standard or obvious, but if it is, I don't know where it's written down: Is it possible to define equivariant cohomology of a topological space in terms of some notion of "equivariant singular cochains", that is, without using the Borel construction?

Answer by Kevin McGerty for Equivariant singular cohomology

2009-12-23T03:02:16Z

Here's an answer which I learned from Goresky-Kottwitz-MacPherson's paper on equivariant cohomology and Koszul duality: they use some notion of geometric chain which is probably something like subanalytic chains, but anyway, the idea is as follows.

Suppose $G$ is a compact Lie group of dimension d. An abstract equivariant $k$-chain $c$ is a $(k+d)$-dimensional chain in some $\mathbb R^n$ (or perhaps it's better to say $\mathbb R^\infty$) equipped with a free action of $G$. Then if $X$ is a $G$-space, an equivariant chain in $X$ is a $K$-equivariant map from an abstract chain to $X$. You can obviously form a chain complex out of these things, and the result gives you the equivariant cohomology of $X$ (in the Borel construction say).

Answer by Ben Webster for Equivariant singular cohomology

2009-12-23T03:57:23Z

You can think of it as the cohomology of the simplicial manifold $X\leftleftarrows X\times G \cdots$ where the $n$-simplices are $X\times G^n$ and the face maps either act on $X$ or multiply two consecutive entries.

Of course, some people will tell you that that is really the same as the Borel construction, but if you're willing to interpret things that liberally, you'll never get away from the Borel construction.

Answer by Chris Gerig for Equivariant singular cohomology

2011-02-24T22:44:57Z

This isn't singular, but in the same spirit as what you want (i.e. not the Borel construction):

Take the cellular chain complex $C(X)$ of a $G$-complex $X$. You can define cohomology groups with coefficients in a chain complex, and $H^*(G,C(X))$ is defined to be the equivariant cohomology of $X$ (for $X$ finite-dimensional and $G$ finite).

This is explained in Ken Brown's $\textit{Cohomology of Groups}$ (chapter VII).

Answer by Peter May for Equivariant singular cohomology

2011-02-25T04:35:52Z

In 1965 or so, Glen Bredon defined ordinary equivariant cohomology, ordinary meaning that it satisfies the dimension axiom: For each coefficient system $M$ (contravariant functor from the orbit category of G to the category of Abelian groups), there is a unique cohomology theory $H^*_G(-;M)$ such that, when restricted to the orbit category, it spits out the functor $M$. Just as in the nonequivariant world, it can be defined using either singular or cellular cochains, the latter defined using $G$-CW complexes. This works as stated for any topological group $G$.

For an abelian group $A$, Borel cohomology with coefficients in $A$, $H^*(EG\times_G X;A)$ is the extremely special case in which one takes $M$ to be the constant coefficient system $\underline{A}$ at the group $A$ and replaces $X$ by $EG\times X$. That is,

$$ H^*_G(EG\times X; \underline{A}) = H^*(EG\times_G X;A) $$